A regular triangle is inscribed in a circle. Find the area of the smaller segment of the cut

A regular triangle is inscribed in a circle. Find the area of the smaller segment of the cut off one of the sides of the triangle if the radius of the circle is 18 cm.

Determine the area of ​​the circle. Scr = π * R2 = π * 324 cm2.

The radii ОА, ОВ, OC divide the circle into three equal sectors, then the area of ​​the AOB sector is equal to:

Ssec = Scr / 3 = π * 324/3 = π * 108 cm2.

In triangle AOB, angle AOB = 360/3 = 120. Triangle AOB is isosceles, since OA = OB = 18 cm, then the height OH is also its median and bisector, then angle BOH = 120/2 = 60, angle OBH = (90 – 60) = 30. Then the leg OH lies opposite the angle 30, then OH = OB / 2 = 18/2 = 9 cm.

BH ^ 2 = OB ^ 2 – OH ^ 2 = 324 – 81 = 243.

BH = 9 * √3 cm.Then AB = 2 * BH = 18 * √3 cm.

Determine the area of ​​the triangle ABC.

Savs = AB * OH / 2 = 18 * √3 * 9/2 = 81 * √3 cm2.

The area of ​​the segment is equal to: Sseg = Ssec – Savs = π * 108 – 81 * √3 = 27 * (4 * π – 3 * √3) cm2.

Answer: The area of ​​the segment is 27 * (4 * π – 3 * √3) cm2.



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